Hello again gurus,
assume we sample an analog signal: gi=g(i*dt).

The real energy of the signal is given by the integral of the square
of the signal:
E=integral[ |g(t)|^2*dt ]

However when we use n samples we approximate this integral with a
finite sum:
En=sum[ |gi|^2*dt ]

This sum should approach the integral as n is increasing:
lim n->infinity { En } = E

Since En != E there is an "error" associated with sampling a signal.

Assume we find the frequency spectrum of the signal:
G(f) = DFT[ g(t) ]

1.
How will this "error" appear in the Fourier space?

2.
We all know that when sampling we should follow Nyquist to avoid
aliasing:
dt<=1/(2*fmax)
Should we also study how the energy resulting from our sampling: En
approaches E as a function of n,
and then choose a sampling so large that En/E ~= 1?


Thanks in advance for any answers!
Andreas Werner Paulsen

Andy365 wrote:

> Hello again gurus,

No gurus.
DSP = Dipshits, Stupidents and Posers (c) Jacobsen.

> assume we sample an analog signal: gi=g(i*dt).
> 
> The real energy of the signal is given by the integral of the square
> of the signal:
> E=integral[ |g(t)|^2*dt ]
> 
> However when we use n samples we approximate this integral with a
> finite sum:
> En=sum[ |gi|^2*dt ]
> 
> This sum should approach the integral as n is increasing:
> lim n->infinity { En } = E
> 
> Since En != E there is an "error" associated with sampling a signal.

This is what the aliasing error is about.
Finite n -> finite length -> infinite spectrum -> aliasing.

> Assume we find the frequency spectrum of the signal:
> G(f) = DFT[ g(t) ]
> 
> 1.
> How will this "error" appear in the Fourier space?

As the classic set of the Fourier artifacts due to the finite length of 
the transform window.

> 2.
> We all know that when sampling we should follow Nyquist to avoid
> aliasing:
> dt<=1/(2*fmax)

I don't know what you all know.

> Should we also study how the energy resulting from our sampling: En
> approaches E as a function of n,
> and then choose a sampling so large that En/E ~= 1?

It depends on what do you have, what do you need, and what are you 
trying to do.

> 
> Thanks in advance for any answers!
> Andreas Werner Paulsen

Stupident.



VLV

On Mar 24, 11:14=A0am, Vladimir Vassilevsky <nos...@nowhere.com> wrote:
> Andy365 wrote:
> > Hello again gurus,
>
> No gurus.
> DSP =3D Dipshits, Stupidents and Posers (c) Jacobsen.
>
....
>
> Stupident.

i guess i'm a poser.

which one are you, Vlad?

r b-j

On Mar 24, 10:45=A0am, Andy365 <andreas_w_paul...@yahoo.com> wrote:
> Hello again gurus,

"Ooooommmm..."

> assume we sample an analog signal: gi=3Dg(i*dt).
>
> The real energy of the signal is given by the integral of the square
> of the signal:
> E=3Dintegral[ |g(t)|^2*dt ]
>
> However when we use n samples we approximate this integral with a
> finite sum:
> En=3Dsum[ |gi|^2*dt ]
>
> This sum should approach the integral as n is increasing:
> lim n->infinity { En } =3D E

on thing that you have to think about a little is dimensional
analysis.  you must compare apples to apples and your En is in terms
of volts^2 (assuming g_i or g(t) is in volts) and E is in volt^2 *
time.  not the same species of animal.


> Since En !=3D E there is an "error" associated with sampling a signal.
>
> Assume we find the frequency spectrum of the signal:
> G(f) =3D DFT[ g(t) ]
>

i don't think that the *discrete* FT operates on the continuous-time
g(t).

> 1.
> How will this "error" appear in the Fourier space?
>
> 2.
> We all know that when sampling we should follow Nyquist to avoid
> aliasing:
> dt<=3D1/(2*fmax)
> Should we also study how the energy resulting from our sampling: En
> approaches E as a function of n,
> and then choose a sampling so large that En/E ~=3D 1?
>
> Thanks in advance for any answers!
> Andreas Werner Paulsen

On Mar 24, 4:14=A0pm, Vladimir Vassilevsky <nos...@nowhere.com> wrote:
> Andy365 wrote:
> > Hello again gurus,
>
> No gurus.
> DSP =3D Dipshits, Stupidents and Posers (c) Jacobsen.
>
> > assume we sample an analog signal: gi=3Dg(i*dt).
>
> > The real energy of the signal is given by the integral of the square
> > of the signal:
> > E=3Dintegral[ |g(t)|^2*dt ]
>
> > However when we use n samples we approximate this integral with a
> > finite sum:
> > En=3Dsum[ |gi|^2*dt ]
>
> > This sum should approach the integral as n is increasing:
> > lim n->infinity { En } =3D E
>
> > Since En !=3D E there is an "error" associated with sampling a signal.
>
> This is what the aliasing error is about.
> Finite n -> finite length -> infinite spectrum -> aliasing.
>
> > Assume we find the frequency spectrum of the signal:
> > G(f) =3D DFT[ g(t) ]
>
> > 1.
> > How will this "error" appear in the Fourier space?
>
> As the classic set of the Fourier artifacts due to the finite length of
> the transform window.
>
> > 2.
> > We all know that when sampling we should follow Nyquist to avoid
> > aliasing:
> > dt<=3D1/(2*fmax)
>
> I don't know what you all know.
>
> > Should we also study how the energy resulting from our sampling: En
> > approaches E as a function of n,
> > and then choose a sampling so large that En/E ~=3D 1?
>
> It depends on what do you have, what do you need, and what are you
> trying to do.
>
>
>
> > Thanks in advance for any answers!
> > Andreas Werner Paulsen
>
> Stupident.
>
> VLV

Thank you VLV for your answer!
So you are saying that the difference in energy between the continous
and the discrete case is the same effect as aliasing?
A practical problem with Nyquist is that we need to know fmax
beforehand (e.g. maybe the signal was lowpassfiltered to fmax by an
analog process).
If I have an analog signal with an unknown fmax; could I not then use
a plot of En(n) to determine an appropriate n?
As n -> infinity En would approach E. From such a plot I may e.g.
conclude that n=3D1024 samples would be sufficient since E1024 seemed to
be very close to the asymptotical E.

Andreas W. P.

On Mar 24, 4:28=A0pm, robert bristow-johnson <r...@audioimagination.com>
wrote:
> On Mar 24, 10:45=A0am, Andy365 <andreas_w_paul...@yahoo.com> wrote:
>
> > Hello again gurus,
>
> "Ooooommmm..."
>
> > assume we sample an analog signal: gi=3Dg(i*dt).
>
> > The real energy of the signal is given by the integral of the square
> > of the signal:
> > E=3Dintegral[ |g(t)|^2*dt ]
>
> > However when we use n samples we approximate this integral with a
> > finite sum:
> > En=3Dsum[ |gi|^2*dt ]
>
> > This sum should approach the integral as n is increasing:
> > lim n->infinity { En } =3D E
>
> on thing that you have to think about a little is dimensional
> analysis. =A0you must compare apples to apples and your En is in terms
> of volts^2 (assuming g_i or g(t) is in volts) and E is in volt^2 *
> time. =A0not the same species of animal.
>
> > Since En !=3D E there is an "error" associated with sampling a signal.
>
> > Assume we find the frequency spectrum of the signal:
> > G(f) =3D DFT[ g(t) ]
>
> i don't think that the *discrete* FT operates on the continuous-time
> g(t).
>
>
>
> > 1.
> > How will this "error" appear in the Fourier space?
>
> > 2.
> > We all know that when sampling we should follow Nyquist to avoid
> > aliasing:
> > dt<=3D1/(2*fmax)
> > Should we also study how the energy resulting from our sampling: En
> > approaches E as a function of n,
> > and then choose a sampling so large that En/E ~=3D 1?
>
> > Thanks in advance for any answers!
> > Andreas Werner Paulsen- Hide quoted text -
>
> - Show quoted text -

Sorry Robert,
should be
Gi=3DG(fi) =3D DFT[ gi=3Dg(ti) ]

When it comes to dimension:
no both the integral and the sum has the dimension: Volt^2*time.


A.W.P.

On Mar 24, 9:45=A0am, Andy365 <andreas_w_paul...@yahoo.com> wrote:

> The real energy of the signal is given by the integral of the square
> of the signal:
> E=3Dintegral[ |g(t)|^2*dt ]

OK.  The signal is rect(t) which has value 1 for |t| < 0.5
and value 0 for |t| > 0.5.  Thus, it is easy to compute
that E =3D 1.

>
> However when we use n samples we approximate this integral with a
> finite sum:
> En=3Dsum[ |gi|^2*dt ]
>
> This sum should approach the integral as n is increasing:
> lim n->infinity { En } =3D E

OK.  Let's sample rect(t) at n points spaced 0.3 apart, i.e.
at t =3D ...., -0.6, -0.3, 0, 0.3, 0.6, ... etc.  So, En =3D 3
as n increases and so the limit does not approach E =3D 1; the
limit is 3.

Oh, you mean the n points are spaced closer and closer together
as n increases?  Well, in that case, more and more points fall
in the interval (-0.5,0.5) and so En is an increasing function
on n that does not converge to a limit.

So, how exactly are you doing the sampling, and in what sense
does sum[ |gi|^2*dt ], which I assume means sum[ |g(t_i)|^2 ],
approach E?  (Hint: reading about Riemann sums and paying
attention to the details of the Delta's might help here.]

--Dilip Sarwate

On Mar 24, 5:12=A0pm, dvsarwate <dvsarw...@yahoo.com> wrote:
> On Mar 24, 9:45=A0am, Andy365 <andreas_w_paul...@yahoo.com> wrote:
>
> > The real energy of the signal is given by the integral of the square
> > of the signal:
> > E=3Dintegral[ |g(t)|^2*dt ]
>
> OK. =A0The signal is rect(t) which has value 1 for |t| < 0.5
> and value 0 for |t| > 0.5. =A0Thus, it is easy to compute
> that E =3D 1.
>
>
>
> > However when we use n samples we approximate this integral with a
> > finite sum:
> > En=3Dsum[ |gi|^2*dt ]
>
> > This sum should approach the integral as n is increasing:
> > lim n->infinity { En } =3D E
>
> OK. =A0Let's sample rect(t) at n points spaced 0.3 apart, i.e.
> at t =3D ...., -0.6, -0.3, 0, 0.3, 0.6, ... etc. =A0So, En =3D 3
> as n increases and so the limit does not approach E =3D 1; the
> limit is 3.
>
> Oh, you mean the n points are spaced closer and closer together
> as n increases? =A0Well, in that case, more and more points fall
> in the interval (-0.5,0.5) and so En is an increasing function
> on n that does not converge to a limit.
>
> So, how exactly are you doing the sampling, and in what sense
> does sum[ |gi|^2*dt ], which I assume means sum[ |g(t_i)|^2 ],
> approach E? =A0(Hint: reading about Riemann sums and paying
> attention to the details of the Delta's might help here.]
>
> --Dilip Sarwate

Hello Dilip,
in the case of rect(t) the sum goes from t=3D-0.5->0.5 over n samples
so: dt=3D(0.5-(-0.5))/n =3D 1/n.
Therefore:
En =3D sum[ |1|^2*1/n] =3D 1 for every n.

A.W.P.

On Mar 24, 12:38=A0pm, Andy365 <andreas_w_paul...@yahoo.com> wrote:

>
> Hello Dilip,
> in the case of rect(t) the sum goes from t=3D-0.5->0.5 over n samples
> so: dt=3D(0.5-(-0.5))/n =3D 1/n.
> Therefore:
> En =3D sum[ |1|^2*1/n] =3D 1 for every n.
>
> A.W.P.

But n samples have n-1 spaces between them, so dt =3D 1/(n-1)
and so En =3D n/(n-1) and converges to 1 from above. You
really need to pay attention to the details if you are
going to be asking questions like the ones you raised in
your original post, where you didn't bother to define dt,
seemed unaware of the difference between the DFT and the
continuous-time Fourier transform, and so on.

--Dilip Sarwate

On Mar 24, 7:00=A0pm, dvsarwate <dvsarw...@yahoo.com> wrote:
> On Mar 24, 12:38=A0pm, Andy365 <andreas_w_paul...@yahoo.com> wrote:
>
>
>
> > Hello Dilip,
> > in the case of rect(t) the sum goes from t=3D-0.5->0.5 over n samples
> > so: dt=3D(0.5-(-0.5))/n =3D 1/n.
> > Therefore:
> > En =3D sum[ |1|^2*1/n] =3D 1 for every n.
>
> > A.W.P.
>
> But n samples have n-1 spaces between them, so dt =3D 1/(n-1)
> and so En =3D n/(n-1) and converges to 1 from above. You
> really need to pay attention to the details if you are
> going to be asking questions like the ones you raised in
> your original post, where you didn't bother to define dt,
> seemed unaware of the difference between the DFT and the
> continuous-time Fourier transform, and so on.
>
> --Dilip Sarwate

Hello Dilip,
yes you are right.
So from the above E101 =3D 101/100=3D1.01.
This is 1 % more energy than the continous case (E=3D1).
SInce this is a very small "error" can we conclude that choosing n =3D
101 samples would suffice?
How does this "error" relate to aliasing?
To me it seems that Vladimir advocates that these are the same thing.
I do not understand how this can be, since Nyquist specify a sampling
rate above which there will be no aliasing,
whereas the "energy error" (En/E - 1) will never fully vanish.

Andreas Werner Paulsen

On Mar 24, 11:48=A0am, Andy365 <andreas_w_paul...@yahoo.com> wrote:
> On Mar 24, 4:28=A0pm, robert bristow-johnson <r...@audioimagination.com>
> wrote:
>
>
>
> > On Mar 24, 10:45=A0am, Andy365 <andreas_w_paul...@yahoo.com> wrote:
>
> > > Hello again gurus,
>
> > "Ooooommmm..."
>
> > > assume we sample an analog signal: gi=3Dg(i*dt).
>
> > > The real energy of the signal is given by the integral of the square
> > > of the signal:
> > > E=3Dintegral[ |g(t)|^2*dt ]
>
> > > However when we use n samples we approximate this integral with a
> > > finite sum:
> > > En=3Dsum[ |gi|^2*dt ]
>
> > > This sum should approach the integral as n is increasing:
> > > lim n->infinity { En } =3D E
>
> > on thing that you have to think about a little is dimensional
> > analysis. =A0you must compare apples to apples and your En is in terms
> > of volts^2 (assuming g_i or g(t) is in volts) and E is in volt^2 *
> > time. =A0not the same species of animal.
>
> > > Since En !=3D E there is an "error" associated with sampling a signal=
..
>
> > > Assume we find the frequency spectrum of the signal:
> > > G(f) =3D DFT[ g(t) ]
>
> > i don't think that the *discrete* FT operates on the continuous-time
> > g(t).
>
> > > 1.
> > > How will this "error" appear in the Fourier space?
>
> > > 2.
> > > We all know that when sampling we should follow Nyquist to avoid
> > > aliasing:
> > > dt<=3D1/(2*fmax)
> > > Should we also study how the energy resulting from our sampling: En
> > > approaches E as a function of n,
> > > and then choose a sampling so large that En/E ~=3D 1?
>
> > > Thanks in advance for any answers!
> > > Andreas Werner Paulsen- Hide quoted text -
>
> > - Show quoted text -
>
> Sorry Robert,
> should be
> Gi=3DG(fi) =3D DFT[ gi=3Dg(ti) ]
>
> When it comes to dimension:
> no both the integral and the sum has the dimension: Volt^2*time.

well, you made discrete points (that's helpful in defining the DFT)
but you have done nothing about the dimensional difference.

hint:  somewhere in the mix you will need the sampling period (which
is the reciprocal of the sampling frequency) which has dimension of
time.

another hint, g(t) need not have dimension of voltage, i used that for
illustration only.  but, i think (unless you toss in the V_ref of the
A/D converter), g(t) and g[i] have the same dimension (although their
arguments do not).

r b-j

On 03/24/2011 07:45 AM, Andy365 wrote:
> Hello again gurus,

Who?

> assume we sample an analog signal: gi=g(i*dt).
>
> The real energy of the signal is given by the integral of the square
> of the signal:
> E=integral[ |g(t)|^2*dt ]

Actually, the real energy of the signal is unknown.  If the signal is an 
amplitude, and if it is being applied to a load that acts as a pure 
impedance (i.e., if the signal is volts being applied to a known 
resistance) then we can compute the _real_ energy ("real" in this case 
meaning "actual", not "the non-imaginary part of a complex number").

But that's often the case, so we can fudge it and call 'E' the energy of 
the signal.

> However when we use n samples we approximate this integral with a
> finite sum:
> En=sum[ |gi|^2*dt ]

Taking 'dt' as 'delta t', or the sampling interval, which I'll call 'Ts' 
-- yes.  But the term 'dt' from calculus has no meaning here.

So I would recast this as

gi(n) = g(Ts * n),
En = sum {gi^2 * Ts}

> This sum should approach the integral as n is increasing:
> lim n->infinity { En } = E

Not without putting restrictions on n that affect Ts.  I would say 
rather that the sum should approach the integral as the sample rate 
decreases:

lim Ts -> 0 sum oven n {gi(n)^2 * Ts} = E

> Since En != E there is an "error" associated with sampling a signal.

Under what conditions is En != E?  Is this necessarily so for all Ts and 
all g(t)?

> Assume we find the frequency spectrum of the signal:
> G(f) = DFT[ g(t) ]
 >
 > 1.
 > How will this "error" appear in the Fourier space?
 >
 > 2.
 > We all know that when sampling we should follow Nyquist to avoid
 > aliasing:
 > dt<=1/(2*fmax)
 > Should we also study how the energy resulting from our sampling: En
 > approaches E as a function of n,
 > and then choose a sampling so large that En/E ~= 1?

Here we make the jump that beginning students of DSP make: we forget 
that there are four different flavors of Fourier transform, and we use 
the one that gets stressed in practical DSP, to the exclusion of all others.

g(t) is a continuous function of time, and therefore cannot have it's 
DFT taken.  Further, since you haven't put restrictions on it, _I'm_ 
going to assume that g(t) is defined over infinite time.  Thus, we can 
take it's _Fourier transform_:

G(omega) = F{g(t)}

gi(n) is not a continuous function of time -- it's a function of the 
discrete index n.  Thus, we can't take it's (continuous time) Fourier 
transform.  We can, instead, take it's discrete-time Fourier transform. 
  For lack of better notation, I'll also define this as F{gi}:

Gi(w) = F{gi(n)}

Here we run into some notational difficulty: omega and w are different 
critters, existing in different spaces and having different units.  The 
frequency variable omega has units radians/sec, and is defined over the 
entire real line.  Conversely, the frequency variable w has units of 
radians, and can be taken as being defined over any segment of the real 
line that is 2*pi radians in extent, but after that we find that G(w) is 
periodic.

But, _if_ G(omega) is zero for any |omega| >= omega_0,
and _if_ omega_0 <= pi/Ts, then we find (possibly with some scaling 
differences) that Gi(omega * Ts) = G(omega).

Then I note without proof (it'll be in your books) that Parseval's 
theorem works for the discrete-time Fourier transform just as it does 
for the continuous-time Fourier transform.  Thus (again with scaling 
difficulties, because I can never remember where to put my factors of pi 
without reference to books, and I'm lazy):

int from -infinity to infinity (G(omega)^2 domega) =
   K int from -infinity to infinity (g(t)^2 dt)

and

int from -pi to pi (Gi(w)^2 domega) =
   K sum from n = -infinity to infinity {gi(n)^2}

Now you're just a factor of Ts, and maybe the odd 2*pi, from finding out 
that _for a bandlimited signal_, _sampled above Nyquist_, _properly 
scaled_, E = En.

Unless I got my math wrong, and I've done enough free work for one day!

-- 

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

Do you need to implement control loops in software?
"Applied Control Theory for Embedded Systems" was written for you.
See details at http://www.wescottdesign.com/actfes/actfes.html

On Mar 24, 9:46=A0pm, Tim Wescott <t...@seemywebsite.com> wrote:
> On 03/24/2011 07:45 AM, Andy365 wrote:
>
> > Hello again gurus,
>
> Who?
>
> > assume we sample an analog signal: gi=3Dg(i*dt).
>
> > The real energy of the signal is given by the integral of the square
> > of the signal:
> > E=3Dintegral[ |g(t)|^2*dt ]
>
> Actually, the real energy of the signal is unknown. =A0If the signal is a=
n
> amplitude, and if it is being applied to a load that acts as a pure
> impedance (i.e., if the signal is volts being applied to a known
> resistance) then we can compute the _real_ energy ("real" in this case
> meaning "actual", not "the non-imaginary part of a complex number").
>
> But that's often the case, so we can fudge it and call 'E' the energy of
> the signal.
>
> > However when we use n samples we approximate this integral with a
> > finite sum:
> > En=3Dsum[ |gi|^2*dt ]
>
> Taking 'dt' as 'delta t', or the sampling interval, which I'll call 'Ts'
> -- yes. =A0But the term 'dt' from calculus has no meaning here.
>
> So I would recast this as
>
> gi(n) =3D g(Ts * n),
> En =3D sum {gi^2 * Ts}
>
> > This sum should approach the integral as n is increasing:
> > lim n->infinity { En } =3D E
>
> Not without putting restrictions on n that affect Ts. =A0I would say
> rather that the sum should approach the integral as the sample rate
> decreases:
>
> lim Ts -> 0 sum oven n {gi(n)^2 * Ts} =3D E
>
> > Since En !=3D E there is an "error" associated with sampling a signal.
>
> Under what conditions is En !=3D E? =A0Is this necessarily so for all Ts =
and
> all g(t)?
>
> > Assume we find the frequency spectrum of the signal:
> > G(f) =3D DFT[ g(t) ]
>
> =A0>
> =A0> 1.
> =A0> How will this "error" appear in the Fourier space?
> =A0>
> =A0> 2.
> =A0> We all know that when sampling we should follow Nyquist to avoid
> =A0> aliasing:
> =A0> dt<=3D1/(2*fmax)
> =A0> Should we also study how the energy resulting from our sampling: En
> =A0> approaches E as a function of n,
> =A0> and then choose a sampling so large that En/E ~=3D 1?
>
> Here we make the jump that beginning students of DSP make: we forget
> that there are four different flavors of Fourier transform, and we use
> the one that gets stressed in practical DSP, to the exclusion of all othe=
rs.
>
> g(t) is a continuous function of time, and therefore cannot have it's
> DFT taken. =A0Further, since you haven't put restrictions on it, _I'm_
> going to assume that g(t) is defined over infinite time. =A0Thus, we can
> take it's _Fourier transform_:
>
> G(omega) =3D F{g(t)}
>
> gi(n) is not a continuous function of time -- it's a function of the
> discrete index n. =A0Thus, we can't take it's (continuous time) Fourier
> transform. =A0We can, instead, take it's discrete-time Fourier transform.
> =A0 For lack of better notation, I'll also define this as F{gi}:
>
> Gi(w) =3D F{gi(n)}
>
> Here we run into some notational difficulty: omega and w are different
> critters, existing in different spaces and having different units. =A0The
> frequency variable omega has units radians/sec, and is defined over the
> entire real line. =A0Conversely, the frequency variable w has units of
> radians, and can be taken as being defined over any segment of the real
> line that is 2*pi radians in extent, but after that we find that G(w) is
> periodic.
>
> But, _if_ G(omega) is zero for any |omega| >=3D omega_0,
> and _if_ omega_0 <=3D pi/Ts, then we find (possibly with some scaling
> differences) that Gi(omega * Ts) =3D G(omega).
>
> Then I note without proof (it'll be in your books) that Parseval's
> theorem works for the discrete-time Fourier transform just as it does
> for the continuous-time Fourier transform. =A0Thus (again with scaling
> difficulties, because I can never remember where to put my factors of pi
> without reference to books, and I'm lazy):
>
> int from -infinity to infinity (G(omega)^2 domega) =3D
> =A0 =A0K int from -infinity to infinity (g(t)^2 dt)
>
> and
>
> int from -pi to pi (Gi(w)^2 domega) =3D
> =A0 =A0K sum from n =3D -infinity to infinity {gi(n)^2}
>
> Now you're just a factor of Ts, and maybe the odd 2*pi, from finding out
> that _for a bandlimited signal_, _sampled above Nyquist_, _properly
> scaled_, E =3D En.
>
> Unless I got my math wrong, and I've done enough free work for one day!
>
> --
>
> Tim Wescott
> Wescott Design Serviceshttp://www.wescottdesign.com
>
> Do you need to implement control loops in software?
> "Applied Control Theory for Embedded Systems" was written for you.
> See details athttp://www.wescottdesign.com/actfes/actfes.html

Thank you Tim for a very well written answer!
(I wish I had been as precise in my initial post.)

Andreas

You have ignored a basic requirement for a sampled signal. To wit: the samp=
le rate must exceed twice the highest frequency present in the sampled sign=
al. If that requirement is met, there is no loss of information. All that c=
an be learned about the original signal can be learned from the samples.

To be sure, there are circumstances that don't need complete information. S=
ervo systems can gain more from eliminating the delay of a proper anti-alia=
s filter than they lose from the aliasing incurred. The questions you ask a=
ren't appropriate those circumstances.

Jerry
--=20
Engineering is the art of making what you want from things you can get.

On Mar 24, 4:46=A0pm, Tim Wescott <t...@seemywebsite.com> wrote:
....
> Then I note without proof (it'll be in your books) that Parseval's
> theorem works for the discrete-time Fourier transform just as it does
> for the continuous-time Fourier transform. =A0Thus (again with scaling
> difficulties, because I can never remember where to put my factors of pi
> without reference to books, and I'm lazy):

the solution to the problem of remembering when and where to put the
factors of pi or 2*pi is to use the "unitary, ordinary frequency"
convention of the FT:

http://en.wikipedia.org/wiki/Fourier_transform#Other_conventions

you need to remember the 2*pi that goes with "f".  use df and dt.  the
inverse FT is just like the forward FT (the -j or +j is of no
consequence).  there are no external factors for either the
convolution theorem nor Parsevals.  and using "duality" (like what is
the FT of the sinc function?) is trivial.

On Mar 24, 10:45=A0am, Andy365 <andreas_w_paul...@yahoo.com> wrote:
> Hello again gurus,
> assume we sample an analog signal: gi=3Dg(i*dt).
>
> The real energy of the signal is given by the integral of the square
> of the signal:
> E=3Dintegral[ |g(t)|^2*dt ]
>
> However when we use n samples we approximate this integral with a
> finite sum:
> En=3Dsum[ |gi|^2*dt ]
>
> This sum should approach the integral as n is increasing:
> lim n->infinity { En } =3D E
>
> Since En !=3D E there is an "error" associated with sampling a signal.
>
> Assume we find the frequency spectrum of the signal:
> G(f) =3D DFT[ g(t) ]
>
> 1.
> How will this "error" appear in the Fourier space?
>
> 2.
> We all know that when sampling we should follow Nyquist to avoid
> aliasing:
> dt<=3D1/(2*fmax)
> Should we also study how the energy resulting from our sampling: En
> approaches E as a function of n,
> and then choose a sampling so large that En/E ~=3D 1?
>
> Thanks in advance for any answers!
> Andreas Werner Paulsen

Hello Andy,

There are several ways of looking at this. If you assume your signal
were bandlimited before sampling, then you have have all of the data
to find your integral exactly. Think of your signal as being a linear
combination of time shifted sin(x)/x functions. Then write down your
integral of this composite continuous function in terms of the
individual sin(x)/x functions and your integral now becomes a simple
sum. The errors you see stem from the sin(x)/x functions existing to
infinity in both directions on the x axis so if you are finding a
definite integral then your sin(x)/x functions are truncated. Well
these integrals may be precomputed and thus in terms of your sum
become weights. So instead of your weights all being simple
1,1,1,1,1,1, you end up of the weights near the center of your
interval being nearly equal to one and the ones near the edge being
close to 0.5.

IHTH,

Clay

robert bristow-johnson wrote:

> On Mar 24, 11:14 am, Vladimir Vassilevsky <nos...@nowhere.com> wrote:
> 
>>DSP = Dipshits, Stupidents and Posers (c) Jacobsen.
> 
> i guess i'm a poser.
> 
> which one are you, Vlad?

Don't flatter yourself, Robert.


VLV

On 3/24/2011 7:45 AM, Andy365 wrote:
> Hello again gurus,
> assume we sample an analog signal: gi=g(i*dt).
>
> The real energy of the signal is given by the integral of the square
> of the signal:
> E=integral[ |g(t)|^2*dt ]
>
> However when we use n samples we approximate this integral with a
> finite sum:
> En=sum[ |gi|^2*dt ]
>
> This sum should approach the integral as n is increasing:
> lim n->infinity { En } = E
>
> Since En != E there is an "error" associated with sampling a signal.
>
> Assume we find the frequency spectrum of the signal:
> G(f) = DFT[ g(t) ]
>
> 1.
> How will this "error" appear in the Fourier space?
>
> 2.
> We all know that when sampling we should follow Nyquist to avoid
> aliasing:
> dt<=1/(2*fmax)
> Should we also study how the energy resulting from our sampling: En
> approaches E as a function of n,
> and then choose a sampling so large that En/E ~= 1?
>
>
> Thanks in advance for any answers!
> Andreas Werner Paulsen

Andreas,

Great question!  I don't have a "pat" answer but I'll try anyway:

Vlad makes a point re: aliasing but I think the issue is much greater 
that that:

Case I:  Start with infinite-extent continuous time, infinite-extent 
continous frequency.   Then multiply by a regular infinite sequence of 
unit Diracs (impulse functions)  in time.  Depending on the weighting of 
the Diracs, the integral can be held constant before and after.  How you 
get the weights is an exercise for the student I guess (meaning that I 
don't know).  The convention is to weigh them according to the value of 
the function being multiplied.

Case II:  Start with infinite-extent continuous time, finite-extent 
(i.e. absolutely lowpassed) continous frequency.   Then multiply by a 
regular infinite sequence of Diracs (impulse functions)  in time - which 
makes the frequency go periodic.   Shannon helped us figure out that the 
temporal sequence can be perfectly interpolated with sincs centered on 
the samples... which is the same as convolving with a sinc and is the 
same as applying a perfect brick-wall lowpass filter.
As each sinc is weighted by the aligned sample, and because of the 
perfect reconstruction, the result is the same as the original - so 
energy is preserved.  Then you can figure out the relationship between 
the integral of a sinc and this result.

Then (I hope not) one can argue about Diracs, and unit samples, etc. etc.

N doesn't have to increase to go through these steps because it's 
already infinite.

A separate and important question is:

"How did you set the limits of integration in your first equation?"
If the limits of integration were infinite then the result will be 
infinite in most cases.  Reducing the limits of integration reduce the 
"window" in which power is summed.  That doesn't make it wrong at all - 
it's just defined.  So, in the sampled case, increasing N doesn't make 
the energy measure "more correct".  But, it may make the "average power" 
measure "more correct" if you define the objective accordingly.

Fred

On Mar 24, 4:45=A0pm, Andy365 <andreas_w_paul...@yahoo.com> wrote:
> Hello again gurus,
> assume we sample an analog signal: gi=3Dg(i*dt).
>
> The real energy of the signal is given by the integral of the square
> of the signal:
> E=3Dintegral[ |g(t)|^2*dt ]
>
> However when we use n samples we approximate this integral with a
> finite sum:
> En=3Dsum[ |gi|^2*dt ]
>
> This sum should approach the integral as n is increasing:
> lim n->infinity { En } =3D E
>
> Since En !=3D E there is an "error" associated with sampling a signal.
>
> Assume we find the frequency spectrum of the signal:
> G(f) =3D DFT[ g(t) ]
>
> 1.
> How will this "error" appear in the Fourier space?

It doesn't. The DFT is exact, to within the usual issues regarding
numerical round-off, etc.

> 2.
> We all know that when sampling we should follow Nyquist to avoid
> aliasing:

No, we don't. Nyquist told us how to *reduce* aliasing.
One can never remove it completely.

> dt<=3D1/(2*fmax)
> Should we also study how the energy resulting from our sampling: En
> approaches E as a function of n,
> and then choose a sampling so large that En/E ~=3D 1?

No. One should choose a sampling frequency that

1) Allows one to do whatever one likes to do, to the signal
2) Is economically and practically feasible

Prioritized in that order.

Rune